Number Theory
... 3 - if the sum of its digits is divisible by 3. 4 - if the number formed by the last 2 digits is divisible by 4. (ask me why this works) 5 - if the ones digit is 5 or 0. 6 - if it is divisible by 2 AND 3. (All even multiples of 3.) 7 - there is no good trick for 7. 8 - if the number formed by the la ...
... 3 - if the sum of its digits is divisible by 3. 4 - if the number formed by the last 2 digits is divisible by 4. (ask me why this works) 5 - if the ones digit is 5 or 0. 6 - if it is divisible by 2 AND 3. (All even multiples of 3.) 7 - there is no good trick for 7. 8 - if the number formed by the la ...
The Abundancy Index of Divisors of Odd Perfect Numbers
... is an odd perfect number and qi αi ||N ∀i, then the quantity ρi = σ(N/qi αi )/qi αi is an integer (because gcd(qi αi , σ(qi αi )) = 1). Suppose ρi = 1. Then σ(N/qi αi ) = qi αi and σ(qi αi ) = 2N/qi αi . Since N is an odd perfect number, qi is odd, whereupon we have an odd αi by considering parity c ...
... is an odd perfect number and qi αi ||N ∀i, then the quantity ρi = σ(N/qi αi )/qi αi is an integer (because gcd(qi αi , σ(qi αi )) = 1). Suppose ρi = 1. Then σ(N/qi αi ) = qi αi and σ(qi αi ) = 2N/qi αi . Since N is an odd perfect number, qi is odd, whereupon we have an odd αi by considering parity c ...
Elementary Number Theory
... m = pm n = pn1 1 · · · pnr r , 1 · · · pr , where mj , nj are non-negative integers, possibly equal to 0. If we have mj ≤ nj for all j, then m obviously divides n, since the quotient of m/n is a product of primes, hence an integer. The inverse is also true. If m divides n, then n/m has prime factori ...
... m = pm n = pn1 1 · · · pnr r , 1 · · · pr , where mj , nj are non-negative integers, possibly equal to 0. If we have mj ≤ nj for all j, then m obviously divides n, since the quotient of m/n is a product of primes, hence an integer. The inverse is also true. If m divides n, then n/m has prime factori ...
1. Modular arithmetic
... 1)p and q; 2q; ; (p 1)q so we have pq 1 (p 1) (q 1) = (p 1)(q 1) entries. (11) (m n) = (m) (n) whenever GCD(m; n) = 1. Exercise 3.5. Prove the last fact. Exercise 3.6. Compute (1800). Theorem 3.7 (Euler's Formula). If GCD(a; m) = 1, then a m 1 mod m: Exercise 3.8. Compute 17 mod 180 ...
... 1)p and q; 2q; ; (p 1)q so we have pq 1 (p 1) (q 1) = (p 1)(q 1) entries. (11) (m n) = (m) (n) whenever GCD(m; n) = 1. Exercise 3.5. Prove the last fact. Exercise 3.6. Compute (1800). Theorem 3.7 (Euler's Formula). If GCD(a; m) = 1, then a m 1 mod m: Exercise 3.8. Compute 17 mod 180 ...
Class Numbers of Ray Class Fields of Imaginary Quadratic Fields
... used to investigate the integer solutions of polynomials. A well-known example is Fermat’s Last Theorem which states that the equation xp + y p = z p does not have any non-trivial solution for any odd prime p. In 1847, Kummer proved this famous theorem for primes p not dividing the class number of Q ...
... used to investigate the integer solutions of polynomials. A well-known example is Fermat’s Last Theorem which states that the equation xp + y p = z p does not have any non-trivial solution for any odd prime p. In 1847, Kummer proved this famous theorem for primes p not dividing the class number of Q ...
07. Decimals - IntelliChoice.org
... When a fraction is converted to a decimal, it is as either a terminating decimal or a repeating decimal. (a) Fractions representing terminating decimals ...
... When a fraction is converted to a decimal, it is as either a terminating decimal or a repeating decimal. (a) Fractions representing terminating decimals ...
